# Producing a Ricci tensor

1. Apr 10, 2006

### Oxymoron

Producing the Ricci tensor

On a pseudo-Riemannian manifold we can contract the Riemann curvature tensor to form the Ricci tensor. In this process of contraction we sum over two indices to make a (3-1)-tensor into a (2-0)-tensor. My question is, why must we contract two indices? Why cant we contract one index to form a (3-0)-tensor or a (2-1)-tensor, etc...

Last edited: Apr 10, 2006
2. Apr 10, 2006

### Hurkyl

Staff Emeritus
Try it. Use your favorite vector space... say, R². You could use R and it will be even more obvious.

Anyways, try and define a "trace" operation that takes a (1,0)-tensor (that is, a vector in your vector space), and produces a (0,0)-tensor (that is, a scalar).

Oh, and don't forget it should be invariant under change of basis! So...

if Tr(v) = r

then if we apply the change of basis on V given by the transformation A...

then Tr(Av) = r

(because a change of basis doesn't affect scalars)

3. Apr 10, 2006

### Oxymoron

Im not sure what "trace" means, but if I sum the components of my vector I get a scalar. ??

4. Apr 10, 2006

### Hurkyl

Staff Emeritus
Is this operation invariant under a change of basis?

Incidentally, that could be achieved by contracting the tensor product of your vector with a particular covector... so it's nothing new! (And as a benefit, the resulting operation will be invariant under a change of basis... although in other bases it doesn't look like simply adding the components of your vector)

Last edited: Apr 10, 2006
5. Apr 10, 2006

### Oxymoron

I believe that it is not. Changing the basis means the sum will change. In fact, I could change the basis so that the sum is zero, when it was nonzero beforehand.

I may need a hint. :)

6. Apr 10, 2006

### Hurkyl

Staff Emeritus
Well, my example was supposed to make it clear that there isn't a reasonable way to do it.

Also, note that any linear map from V to R is given by simply applying a covector (or equivalently, tensoring with a covector then contracting) -- so we already have all possible such maps in our algebra!

7. Apr 10, 2006

### Oxymoron

Sorry Hurkyl, unfortunately, I'm not seeing your point.

Take the Riemann curvature tensor

$$R(X,Y,Z,\omega)$$

as we can see, it is a (4-1)-tensor with vector fields in the first three slots and one 1-form in the fourth slot. (I believe alternatively we could write this as

$$R(X,Y,Z,V^{\flat})$$

but that is another story)

Anyway, we can write the Ricci tensor as

$$Ric(X,Y) = R(X_a,Y,Z,\omega^a)$$

by summing over the first and fourth indices. If we sum over the first two, we get

$$R(X_a,Y^a,Z,\omega) = 0$$

why? Im not sure. Likewise, we could sum over the last two:

$$R(X,Y,Z_a,\omega^a) = 0$$

and it vanishes again. Again, not sure why.

But, if we sum over the first and fourth say (equivalently we could sum over the second and fourth) we do not get zero, and this time I think I know why - because of the Bianchi identities.

The question I propose: Does contracting over two only work because of the fact that our manifold must have a metric.

8. Apr 10, 2006

### Hurkyl

Staff Emeritus
I was trying to demonstrate the problem with the notion of "contracting one index" in the case where we were simply working with a vector. The only way to get from a vector to a scalar is to apply a covector. (Though it may be disguised as a more complicated expression)

Anyways, I'm having type issues with your notation.

So you've defined the Riemann tensor as a map that eats three vectors and a covector, and produces a scalar.

That's fine -- you can contract over its first and fourth indices to produce a rank 2 covector. But your notation seems troublesome:
because the contraction applies to the tensor and not the arguments. In the abstract index notation, R is given by $R_{abc}{}^d$, and you've contracted to produce $R_{abc}{}^a$.

And besides, if X is a vector, then its index should be a superscript!.

Summing over the first two indices is just wrong: they're both subscripts! You have to first apply the metric to raise an index: $R^a{}_{bc}{}^d = g^{a\mu} R_{\mu bc}{}^d$, and then you can contract.

So, your hunch was right for this case. The metric is the only reason we can contract over two like indices -- because what we are really doing is taking an inner product.

Anyways, isn't this one of the antisymmetries of the Riemann tensor? That $R_{abc}{}^d = -R_{bac}{}^d$? Or in the functional notation, R(X, Y, Z, w) = -R(Y, X, Z, w)?

9. Apr 10, 2006

### Perturbation

Contraction is an operation between covariant and contravariant indices. The Ricci tensor is formed by the self contraction of the Riemann tensor, and since there is only one contravariant (raised) index there's only one possible type of tensor one can get from self contraction of the Riemann tensor, a (0 2) tensor.

$$R^a_{bad}=R_{bd}$$

One could contract on other covariant indices other than the middle one, but this is the only contraction that yields a symmetric (0 2) tensor, and the other possible contractions between contra- and covariant indices can be expressible as this symmetric one.

To get a (0 3) tensor we'd have to supplement a covariant vector on which to operate R on, thus removing this "slot" in the tensor's argument.

$$\omega_aR^a_{bad}$$

Last edited: Apr 10, 2006